A draft

Let us consider a Ring on ${\mathbb{K}}^2$$\mathbb K^2$, where $\mathbb{K}$$\mathbb K$ is algebraically closed.

The operation is defined by $(u,v)+(x,y)=(u+x,v+y),(u,v)\ast (x,y)=(ux+2vy,uy+vx).$$(u,v)+(x,y)=(u+x,v+y),(u,v)*(x,y)=(ux+2vy,uy+vx).$

One question arises: Does $({\mathbb{K}}^2,+,\ast )$$(\mathbb K^2,+,*)$ form an integral domain?

As we can see $(u,v)\ast (v,y)=0⟺ux+2vy=0\text{ and }uy+vx=0$$(u,v)*(v,y)=0\iff ux+2vy=0\text{ and }uy+vx=0$

One interesting way is

Which is equivalent to seeing whether the intersection of these two zero locus forms a nontrivial algebraic variety.

By Hilbert's Nullstellensatz, it is equivalent to asking does $(ux+2vy)\cap (uy+vx)$$(ux+2vy)\cap(uy+vx)$ form a radical ideal in $\mathbb{K}[u,v,x,y]$$\mathbb K[u,v,x,y]$?

If $\mathbb{K}$$\mathbb K$ is a field in general, for example, $\mathbb{R}$$\mathbb R$.

Let $u\ne 0$$u\ne 0$, then $x+2v^{\prime }y=0$$x+2v'y=0$ and $y+v^{\prime }x=0$$y+v'x=0$, hence $-y+2v^{\prime 2}y=0,y(2v^{\prime 2}-1)=0.$$-y+2v'^2y=0,y(2v'^2-1)=0.$ Let $y\ne 0,v^{\prime 2}=1/2$$y\ne 0,v'^2=1/2$

$y+1/\sqrt{2}x=0⟺\sqrt{2}y+x=0$$y+1/\sqrt 2x=0\iff\sqrt 2y+x=0$

Hence $(1,1/\sqrt{2})\ast (x,y)=(x+\sqrt{2}y,y+1/\sqrt{2}x)=(0,0)$$(1,1/\sqrt 2)*(x,y)=(x+\sqrt 2 y,y+1/\sqrt 2 x)=(0,0)$